Verifying a vector identity using Levi-Civita | Quantum Science Philippines
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Verifying a vector identity using Levi-Civita

VERIFYING A VECTOR IDENTITY USING LEVI-CIVITA

Bianca Rae B. Sambo

Hello physics enthusiast! I am BR, a graduate student in Physics in Mindanao State University – Iligan Institute of Technology. Hope my solution will be of use to you. Keep visiting this site!

 

The vector identity to be verified using Levi-Civita is,

 \vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+\vec{a}\times(\vec{\nabla}\times\vec{b})+\vec{b}\times(\vec{\nabla}\times\vec{a}) 

where we define the following vectors as,

 \vec{a} = a_{i} \hat{i}
\vec{b} = b_{j} \hat{j}
 \vec{\nabla} = \partial{k } \hat{k}

We start by expanding the left hand side of the identity.

   \vec{\nabla}(\vec{a}\bullet\vec{b})=\vec{\nabla} (\delta_{ij} a_{i} b_{j})=\partial{k} \hat{k} (\delta_{ij} a_{i} b_{j})

Then we set i = j and apply the distributive property of the partial derivative.

 

Equation 1:

   \vec{\nabla}(\vec{a}\bullet\vec{b})=\partial{k}(a_{j} b_{j}) \hat{k} =a_{j} \partial{k} (b_{j})\hat{k} + b_{j}\partial{k}(a_{j}) \hat{k}

Notice in the right hand side of the vector identity, we have the equations  (\vec{a}\bullet\vec{\nabla})\vec{b} and (\vec{b}\bullet\vec{\nabla})\vec{a}. These two equations can be written as,

(\vec{a}\bullet\vec{\nabla})\vec{b}= a_{k} \partial{k}(b_{j})\hat{j}

and

(\vec{b}\bullet\vec{\nabla})\vec{a}= b_{k}\partial{k}(a_{i}) \hat{i}

 

So we incorporate this to Equation 1 above.

   \vec{\nabla}(\vec{a}\bullet\vec{b})= a_{k} \partial{k}(b_{j}) \hat{j}+ b_{k}\partial{k}(a_{i}) \hat{i}+ a_{j}\partial{k} (b_{j}) \hat{k}+b_{j}\partial{k}(a_{j})\hat{k} - a_{k}\partial{k} (b_{j}) \hat{j}- b_{k} \partial{k} (a_{i}) \hat{i}

Equation 2:

   \vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+[a_{j}\partial{k}(b_{j})\hat{k}-a_{k}\partial{k}(b_{j})\hat{j}]+[(b_{j} \partial{k}(a_{j}) \hat{k}- b_{k}\partial{k}(a_{i})\hat{i})]

Notice also that:

   \vec{a}\times(\vec{\nabla}\times\vec{b})=a_{i}\hat{i}\times(\epsilon_{kjm}\partial{k}(b_{j})\hat{m})=\epsilon_{iml}\epsilon_{kjm}a_{i} \partial{k}(b_{j}) \hat{l}=\epsilon_{lim}\epsilon_{kjm}a_{i}\partial{k}(b_{j})\hat{l}=(\delta_{lk}\delta_{ij}-\delta_{lj}\delta_{ik})a_{i}\partial{k}(b_{j})\hat{l}      \vec{a}\times (\vec{\nabla}\times\vec{b})=[a_{j}\partial{k}(b_{j})\hat{k}-a_{k}\partial{k}(b_{j})\hat{j}]

Similarly,

   \vec{b}\times(\vec{\nabla}\times\vec{a})=b_{j}\hat{j}\times(\epsilon_{kin}\partial{k}(a_{i})\hat{n})=\epsilon_{jnp}\epsilon_{kin}b_{j}\partial{k}(a_{i})\hat{p}=\epsilon_{pjn}\epsilon_{kin}b_{j}\partial{k}(a_{i})\hat{p}=(\delta_{pk}\delta_{ji}-\delta_{pi}\delta_{jk}) b_{j}\partial{k}(a_{i})\hat{p}     \vec{b}\times(\vec{\nabla}\times\vec{a})=[b_{j}\partial{k}(a_{j})\hat{k}- b_{k}\partial{k}(a_{i})\hat{i}]
So plugging these equations to equation 2, we finally verified the vector identity because we get:

 \vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+\vec{a}\times(\vec{\nabla}\times\vec{b})+\vec{b}\times(\vec{\nabla}\times\vec{a}) 

 

 

 

 

 

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